1

Math 150

7.1/7.2 – The Law of Sines

2

Q: We know how to solve right triangles using trig, but how can we use trig to solve any triangle?

A: The Law of Sines (7.1/7.2) and The Law of Cosines (7.3).

3

Q: We know how to solve right triangles using trig, but how can we use trig to solve any triangle?

A: The Law of Sines (7.1/7.2) and The Law of Cosines (7.3).

4

The Law of Sines

(also written )

5

Ex 1.Solve triangle if , , and m.

6

Ex 2.The bearing of a lighthouse from a ship was found to be . After the ship sailed 5.8 km due south, the new bearing was . Find the distance between the ship and the lighthouse at each location.

7

To solve a triangle, we’ve needed to know some of its sides and angles. We can classify the possibilities like this (A = Angle, S = Side):

Case 1: ASA or SAA

8

Case 2: SSA

9

Case 3: SAS

10

Case 4: SSS

11

Note: The Law of Sines helps us solve cases 1 and 2. The Law of Cosines helps us solve cases 3 and 4.

12

What if we’re given the following info and asked to solve the triangle? This is Case 2 (SSA).But if we try todraw the triangle we get…

Side is too short! We can see this with the Law of Sines.

Pain in the SSA

13

What if we’re given the following info and asked to solve the triangle? This is Case 2 (SSA).But if we try todraw the triangle we get…

Side is too short! We can see this with the Law of Sines.

Pain in the SSA

14

What if we’re given the following info and asked to solve the triangle? This is Case 2 (SSA).But if we try todraw the triangle we get…

Side is too short! We can see this with the Law of Sines.

Pain in the SSA

15

What if we’re given the following info and asked to solve the triangle? This is Case 2 (SSA).But if we try todraw the triangle we get…

Side is too short! We can see this with the Law of Sines.

Pain in the SSA

16

If we make side longer, then we can make a triangle:

Now using the Law of Sines we get…

17

You can imagine if side is in between (and just right), we might get two possible triangles:

This is called the ambiguous case, and it only happens with Case 2 (SSA).

18

You can imagine if side is in between (and just right), we might get two possible triangles:

This is called the ambiguous case, and it only happens with Case 2 (SSA).

19

You can imagine if side is in between (and just right), we might get two possible triangles:

This is called the ambiguous case, and it only happens with Case 2 (SSA).

20

You can imagine if side is in between (and just right), we might get two possible triangles:

This is called the ambiguous case, and it only happens with Case 2 (SSA).

21

Ex 3.Solve the triangle if it exists.

22

Note that and are the two sides that make the angle . Also note that the area formula could be written: or

Area of a Triangle (SAS)

23

Note that and are the two sides that make the angle . Also note that the area formula could be written: or

Area of a Triangle (SAS)

24

Note that and are the two sides that make the angle . Also note that the area formula could be written: or

Area of a Triangle (SAS)

25

Note that and are the two sides that make the angle . Also note that the area formula could be written: or

Area of a Triangle (SAS)

26

Proof that :We can drop a perpendicular down to the base (see right). So, the area is: But so that .Replacing with , we get: QED

27

Proof that :We can drop a perpendicular down to the base (see right). So, the area is: But so that .Replacing with , we get: QED

28

Proof that :We can drop a perpendicular down to the base (see right). So, the area is: But so that .Replacing with , we get: QED

29

Proof that :We can drop a perpendicular down to the base (see right). So, the area is: But so that .Replacing with , we get: QED

30

Proof that :We can drop a perpendicular down to the base (see right). So, the area is: But so that .Replacing with , we get: QED

31

Let’s go back and prove the Law of Sines as well…

32

Proof that :

Using the triangle to the right, we get: and

So, and Thus, Dividing, we get, By making perpendicular segments from and , and using similar reasoning, we get and . Thus, .QED

33

Proof that :

Using the triangle to the right, we get: and

So, and Thus, Dividing, we get, By making perpendicular segments from and , and using similar reasoning, we get and . Thus, .QED

34

Proof that :

Using the triangle to the right, we get: and

So, and Thus, Dividing, we get, By making perpendicular segments from and , and using similar reasoning, we get and . Thus, .QED

35

Proof that :

Using the triangle to the right, we get: and

So, and Thus, Dividing, we get, By making perpendicular segments from and , and using similar reasoning, we get and . Thus, .QED

36

Proof that :

Using the triangle to the right, we get: and

So, and Thus, Dividing, we get, By making perpendicular segments from and , and using similar reasoning, we get and . Thus, .QED

37

Proof that :

Using the triangle to the right, we get: and

So, and Thus, Dividing, we get, By making perpendicular segments from and , and using similar reasoning, we get and . Thus, .QED

38

Proof that :

Using the triangle to the right, we get: and

So, and Thus, Dividing, we get, By making perpendicular segments from and , and using similar reasoning, we get and . Thus, .QED

39

Proof that :

Using the triangle to the right, we get: and

So, and Thus, Dividing, we get, By making perpendicular segments from and , and using similar reasoning, we get and . Thus, .QED